From a0d7ea15b31e6058c0d7771517aa7eb37801cbb4 Mon Sep 17 00:00:00 2001
From: Erik Strand <erik.strand@cba.mit.edu>
Date: Thu, 2 May 2019 01:10:54 -0400
Subject: [PATCH] Answer 14.1

---
 _psets/11.md | 36 ++++++++++++++++++++++++++++++++++++
 1 file changed, 36 insertions(+)

diff --git a/_psets/11.md b/_psets/11.md
index 7803f38..c088810 100644
--- a/_psets/11.md
+++ b/_psets/11.md
@@ -8,6 +8,42 @@ title: Problem Set 11
 {:.question}
 Do a Taylor expansion of equation (14.6) around V = 0.
 
+Equation 14.6 states
+
+$$
+E \approx 2 E_F - 2 E_C e^{-2/(N_F V)}
+$$
+
+The Taylor expansion of $$e^x$$ about zero is
+
+$$
+e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}
+$$
+
+so
+
+$$
+E \approx 2 E_F - 2 E_C \sum_{n = 0}^\infty \frac{1}{n!} \left( \frac{-2}{N_F V} \right)^n
+$$
+
+But this is an expansion around $$V = \infty$$.
+
+If you really want an expansion about $$V = 0$$, note that
+
+$$
+\frac{d}{dx} e^{x^{-1}} = -x^{-2} e^{x^{-1}}
+$$
+
+As we keep taking higher derivatives we'll get more and more negative powers of $$x$$, but we'll
+never get rid of the $$e^{1/x}$$. So at $$x = 0$$ the latter term dominates, meaning the function
+and all its derivatives are zero. Thus the Taylor expansion about zero is identically zero.
+
+Plugging this into equation 14.6 gives us the rather uninteresting
+
+$$
+E \approx 2 E_F
+$$
+
 
 ## (14.2)
 
-- 
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